Thermal conductivity is commonly denoted as $k$ or $\lambda$. The units $\frac{BTU}{s\cdot in^2 \cdot \frac{^oF}{in}}$you stating are perfectly OK, although awkward. I would have preferred to -at least- simplify the inches:

$$\frac{BTU}{s\cdot in \cdot ^oF}$$

or even better write as:

$$\frac{\frac{BTU}{s}}{ in \cdot ^oF} = \frac{1055\; W}{ in \cdot ^oF}$$

The conversion to SI is the following :

$$1\frac{BTU}{s\cdot in^2 \cdot \frac{^oF}{in}} = 74767.7 \frac{W}{m\cdot K}$$

So the value you've got 9.8 x 10⁻⁶$\frac{BTU}{s\cdot in^2 \cdot \frac{^oF}{in}} $ is equal to approximately $0.7327\frac{W}{m\cdot K}$.

Thermal conductivity is commonly denoted as $k$ or $\lambda$. The units $\mathrm{\frac{BTU}{s\cdot in^2 \cdot \frac{^\circ F}{in}}}$ you state are perfectly OK, although awkward. I would have preferred to—at least—simplify the inches:

$$\mathrm{\frac{BTU}{s\cdot in \cdot ^\circ\mkern-5mu F}}$$

or, even better, write the expression as:

$$\mathrm{\frac{\frac{BTU}{s}}{ in \cdot ^\circ\mkern-5muF} = \frac{1055\; W}{ in \cdot ^\circ\mkern-5muF}}$$

The conversion to SI is as follows:

$$1\mathrm{\frac{BTU}{s\cdot in^2 \cdot \frac{^\circ F}{in}} = 74767.7 \frac{W}{m\cdot K}}$$

So the value you've got of $9.8 \times 10^{-6}\mathrm{\frac{BTU}{s\cdot in^2 \cdot \frac{^\circ F}{in}} }$ is equal to approximately $0.7327\mathrm{\frac{W}{m\cdot K}}$.