Total number of  $\mathrm{cis}-\mathrm{N}-\mathrm{Mn}-\mathrm{Cl}$ Bond angles (that is $\text{Mn}–\text{N and Mn}–\text{Cl}$ Bonds in $cis$ Positions) present in a molecule of $cis-\left[\mathrm{Mn}{\left(\mathrm{en}\right)}_2{\mathrm{Cl}}_2\right]$  The complex is $\left(\mathrm{en}={\mathrm{NH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{NH}}_2\right)$

Number of $cisN-Mn-Cl$ , i.e. $Mn-N$  and  $Mn-Cl$ bonds are in adjacent positions

$1-Cl\left(a\right)-Mn-N\left(1\right) 2-Cl\left(a\right)-Mn-N\left(2\right)$

$3-Cl\left(a\right)-Mn-N\left(4\right) 4-Cl\left(b\right)-Mn-N\left(4\right)$

$5-Cl\left(b\right)-Mn-N\left(3\right) 6-Cl\left(b\right)-Mn-N\left(1\right)$

Hence, the answer is 6.
Note:  All Adjacent positions $\left(at90°\right)$  are $cis$  to each other