Answer
a)
The apparent stellar magnitude of the sun is approximately $-26.9$.
b)
The received intensity of light is approximately $1.58\times {{10}^{-17}}\ \text{W/}{{\text{m}}^{2}}$.
Work Step by Step
a)
$m\left( I \right)=-\left( 19+2.5\cdot \log I \right)$ …… (1)
Now, since, the intensity of sun light is $1390\ \text{W/}{{\text{m}}^{2}}$, substitute $1390\ \text{W/}{{\text{m}}^{2}}$ in equation-1:
$\begin{align}
& m\left( I \right)=-\left( 19+2.5\cdot \log I \right) \\
& m\left( 1390 \right)=-\left( 19+2.5\cdot \log \left( 1390 \right) \right) \\
& m=-\left( 19+2.5\cdot \left( 3.143 \right) \right) \\
& m=-\left( 19+7.857 \right)
\end{align}$
Further simplified,
$\begin{align}
& m=-\left( 19+7.857 \right) \\
& m=-26.857 \\
& m\approx -26.9
\end{align}$
b)
$m\left( I \right)=-\left( 19+2.5\cdot \log I \right)$ …… (1)
$\begin{align}
& m\left( I \right)=-\left( 19+2.5\cdot \log I \right) \\
& 23=-\left( 19+2.5\cdot \log I \right) \\
& -23=19+2.5\cdot \log I \\
& -42=2.5\cdot \log I
\end{align}$
Further simplified,
$\begin{align}
& -42=2.5\cdot \log I \\
& \frac{-42}{2.5}=\log I \\
& -16.8=\log I \\
& {{10}^{-16.8}}=I
\end{align}$
Further simplified,
$\begin{align}
& {{10}^{-16.8}}=I \\
& 1.58\times {{10}^{-17}}\approx I
\end{align}$
