Answer
See below
Work Step by Step
Consider $A=\begin{bmatrix}
a & b\\
c & d\\
e & f
\end{bmatrix}$
We have
$Ker (T)=\{A \in M_{24}(R):T(A)=0\} \\
=\{A \in M_{24}(R):A^T=0\} \\
=\begin{bmatrix}
a & b\\
c & d\\
e & f
\end{bmatrix}\begin{bmatrix}
a & b & c\\
d & e & f
\end{bmatrix}=\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0
\end{bmatrix}; a,b,c,d,e,f \in R\} \\
\rightarrow a=b=c=d=e=f=0$
Hence, $Ker (T)=\{0\}\\
dim[Ker(T)]=0$
According to Rank- Nullity Theorem:
$dim[Ker(T)]+dim[Rng(T)]=dimM_{24}R\\
0+dim[Rng(T)]=6\\
dim[Rng(T)]=6$
Since $Rng(T) \subset M_{42}R$ and $dimM_{42}=6=dim[Rng(T)]$, $Rng(T)=M_{42}R$
