1 $\mathrm{Mn}>\mathrm{Cr}>\mathrm{Ti}>\mathrm{V}$
2 Ti $>\mathrm{V}>\mathrm{Cr}>\mathrm{Mn}$
3 $\mathrm{Cr}>\mathrm{Mn}>\mathrm{V}>\mathrm{Ti}$
4 $\mathrm{V}>\mathrm{Mn}>\mathrm{Cr}>\mathrm{Ti}$
Explanation:
The electronic configuration are follow as-
$$
\begin{aligned}
& \mathrm{Cr}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 3 \mathrm{p}^6, 4 \mathrm{~s}^1, 3 \mathrm{~d}^5 \\
& \mathrm{Mn}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 3 \mathrm{p}^6, 4 \mathrm{~s}^2, 3 \mathrm{~d}^5 \\
& \mathrm{~V}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 3 \mathrm{p}^6, 4 \mathrm{~s}^2, 3 \mathrm{~d}^3 \\
& \mathrm{Ti}-1 \mathrm{~s}^2, 2 \mathrm{~s}^2, 2 \mathrm{p}^6, 3 \mathrm{~s}^2, 3 \mathrm{p}^6, 4 \mathrm{~s}^2, 3 \mathrm{~d}^2
\end{aligned}
$$
The value of the second ionisation energy of $\mathrm{Cr}$ is more due to $\mathrm{Cr}^{+}$has $3 \mathrm{~d}^5$ configuration which is extra energy is required than the $\mathrm{Mn}$.
So, the order of $2^{\text {nd }}$ ionisation energy is-
$$
\mathrm{Cr}>\mathrm{Mn}>\mathrm{V}>\mathrm{Ti}
$$