Considering that the following is already sufficient to solve the "intensive" problem:
Air in the cylinder of a Diesel Engine is at $30\,{}^o\mathrm{C}$ and $138\,\mathrm{kPa}$. If it is further compressed to $1/18\,$th of its original volume, calculate the work done on the system.
the compression ratio is $\varrho=v_1/v_2=18$, and the barometric ratio $\beta=P_2/P_1$ is rapidly derived from the isoentropic relation (given that $\gamma=1.4$ is temperature independent):
$$
P_1v_1^\gamma=P_2v_2^\gamma\to \frac{P_2}{P_1}=\left(\frac{v_1}{v_2}\right)^\gamma\to \beta=\varrho^\gamma
$$
this can be directly used inside the work equation. The latter is derived from the integral of $P\,\mathrm{d}v$ on Clapeyron's Diagram (isoentropic shaft work on a closed system), using the intensive relation, as demonstrated below:
\begin{align}
w=-\int_{v_1}^{v_2}P\,\mathrm{d}v&=-\int_{v_1}^{v_2}\frac{P_1v_1^{\gamma}}{v^\gamma}\,\mathrm{d}v\\
&=P_1v_1^{\gamma}\int_{v_1}^{v_2}-v^{-\gamma}\,\mathrm{d}v\\
&=P_1v_1^{\gamma}\frac{1}{\gamma-1}\left(v_2^{1-\gamma}-v_1^{1-\gamma}\right)\\
&=RT_1\frac{1}{\gamma-1}\left(v_2^{1-\gamma}/v_1^{1-\gamma}-1\right)\\
&=RT_1\frac{1}{\gamma-1}\left(\varrho^{\gamma-1}-1\right)
\end{align}
Since the result will describe an intensive quantity, the extensive result will depend on the chemical substance contained (and not consumed) inside the system. That is derived from the displaced volume and the compression ratio, namely rearranging the terms to highlight $\varrho$:
the compression work done, given the displacement volume of the cylinder as $\Delta V=14.2\,\mathrm{L}$
$$
\Delta V=V_1-V_2\to\frac{\Delta V}{V_2}=\frac{V_1-V_2}{V_2}=\varrho-1\to V_2=\frac{\Delta V}{\varrho-1}
$$
from here the initial volume is known (substituting back the $V_2$ term):
$$
V_1=V_2+\Delta V=\frac{\Delta V}{\varrho-1}+\Delta V=\frac{14.2\,\mathrm{L}}{17}+14.2\,\mathrm{L}=15.03\,\mathrm{L}=15.03\times10^{-3}\,\mathrm{m^3}
$$
substituting all the first intensive parameters inside the Ideal Gas EoS, the constant molar quantity of air is known:
$$
n=\frac{P_1V_1}{RT_1}=\frac{15.03\times10^{-3}\,\mathrm{m^3}\cdot138\times10^{3}\,\mathrm{Pa}}{8.314\,(\mathrm{Pa\,m^3\,mol^{-1}\,K^{-1}})\cdot303.15\,\mathrm{K}}=0.823\,\mathrm{mol}
$$
Knowing that the extensive parameter $W$ is expressed as $nw$, and from the previous equation $w=13.72\,\mathrm{kJ\,mol^{-1}}$, the requested value is $W=0.823\,\mathrm{mol}\cdot 13.72\,\mathrm{kJ\,mol^{-1}}=11.29\,\mathrm{kJ}$.
At this point, even if the result is different from the first given, there can be one explanation. Since in your first method you considered an initial mass of air equal to $1\,\mathrm{kg}$, this is contradictory with respect to the Equation of State. Because the initial (and constant) molar quantity of air depends to the initial extensive state $(T_1,P_1,V_1)$, then the only variable left $V_1$ is derived if the displacement volume and compression ratio are known.
