$(d)$   Since 

$A.\left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right]$ is a scalar matrix and $\left| {3A} \right| = 108$

Suppose the scalar matrix is $\left[ {\begin{array}{*{20}{c}}
k&0\\
0&k
\end{array}} \right]$

$\therefore A.\left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
k&0\\
0&k
\end{array}} \right]$

$ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
k&0\\
0&k
\end{array}} \right]{\left[ {\begin{array}{*{20}{c}}
1&2\\
0&3
\end{array}} \right]^{ – 1}}$

$\left[ {\therefore AB = C \Rightarrow AB{B^{ – 1}} = C{B^{ – 1}} \Rightarrow A = C{B^{ – 1}}} \right]$

$ \Rightarrow A = \frac{1}{3}\left[ {\begin{array}{*{20}{c}}
k&0\\
0&k
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
3&{ – 2}\\
0&1
\end{array}} \right]$

$ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
k&0\\
0&k
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1&{ – \frac{2}{3}}\\
0&{\frac{1}{3}}
\end{array}} \right]$

$ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
k&{ – \frac{2}{3}k}\\
0&{\frac{k}{3}}
\end{array}} \right]\,\,\,\,\,\,\,\,\,…….\left( 1 \right)$

$\because$ $\left| {3A} \right| = 108$

$ \Rightarrow 108 = \left| {\begin{array}{*{20}{c}}
{3k}&{ – 2k}\\
0&k
\end{array}} \right|$

$ \Rightarrow 3{k^2} = 108 \Rightarrow {k^2} = 36 \Rightarrow k =  \pm 6$ 

For $k=6$

$A = \left[ {\begin{array}{*{20}{c}}
6&{ – 4}\\
0&2
\end{array}} \right]$             ….From $(1)$

$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
{36}&{ – 32}\\
0&4
\end{array}} \right]$

For $k=-6$

$ \Rightarrow A = \left[ {\begin{array}{*{20}{c}}
{ – 6}&4\\
0&{ – 2}
\end{array}} \right]$       ….From$(1)$

$ \Rightarrow {A^2} = \left[ {\begin{array}{*{20}{c}}
{36}&{ – 32}\\
0&4
\end{array}} \right]$