Answer
a) $1.84\times10^8\;\rm ly$
b) It is about 1800 times the size of the Milkey Way galaxy.
Work Step by Step
a) Since the intensity is inversely proportional to the distance,
$$\dfrac{I_{Star}}{I_{Sun}}=\dfrac{r^2_{\rm sun \;to\;Earth}}{r^2_{\rm star\;to\;Earth}}$$
Solving for $r_{\rm star\;to\;Earth}$;
$$r_{\rm star\;to\;Earth}^2 =\dfrac{I_{Sun}r^2_{\rm sun \;to\;Earth}}{I_{Star}}$$
Taking the square root for both sides;
$$r_{\rm star\;to\;Earth} =r_{\rm sun \;to\;Earth}\sqrt{\dfrac{I_{Sun}}{I_{Star}}}$$
Plugging the known;
$$r_{\rm star\;to\;Earth} =1.496\times10^{11}\cdot \sqrt{\dfrac{1350}{10^{-23}}}=\bf1.739\times10^{24}\;\rm m$$
Now we need to convert this distance to lightyear.
$$r_{\rm star\;to\;Earth} = 1.739\times10^{24}\;\rm m\cdot \dfrac{1\;ly}{3\times10^8\times 365\times24\times60^2\;m}$$
$$r_{\rm star\;to\;Earth} =\color{red}{\bf 1.84\times10^8}\;\rm ly$$
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b)
To compare these two distances,
$$\dfrac{r_{\rm star\;to\;Earth}}{S_{\rm MW\;galaxy}}=\dfrac{1.84\times10^8}{100\;000}=1838$$
$S$ for size.
Thus,
$$r_{\rm star\;to\;Earth}=\color{red}{\bf1838}\;S_{\rm MW\;galaxy}$$
It is about 1800 times the size of the Milkey Way galaxy.
