In chromyl chloride (${\mathrm{CrO}}_2{\mathrm{Cl}}_2$), the oxidation state of chromium $\left(\mathrm{Cr}\right)$ is $+6.$

So, $\mathrm{Cr} \left(\mathrm{VI}\right) = [\mathrm{Ar}{]}^{18} 3{\mathrm{d}}^0$

For the given elements the configuration is,$\mathrm{Mn}\left(\mathrm{VII}\right) = [\mathrm{Ar}{]}^{18} 3{\mathrm{d}}^0$

$\mathrm{Fe}\left(\mathrm{III}\right) = [\mathrm{Ar}{]}^{18 }3{\mathrm{d}}^5$

$\mathrm{Ti}\left(\mathrm{III}\right) = [\mathrm{Ar}{]}^{18} 3{\mathrm{d}}^1$

$\mathrm{V}\left(\mathrm{IV}\right) = [\mathrm{Ar}{]}^{18} 3{\mathrm{d}}^1$

So among the given options $\mathrm{Mn}\left(\mathrm{VII}\right)$ and $\mathrm{Cr} \left(\mathrm{VI}\right)$ will have same number of d-electrons.