Answer
$C_M$ = 1.412 M
Work Step by Step
${C_{M}}_f$ = $\frac{n_{f}}{V_f}$
First, we need to determine the number of moles of KMn$O_{4}$ in each solution
n = $C_{M}$ $\times$ V
$n_{1}$ = 1.66 $\times$ 0.0352 = 0.058432 moles KMn$O_{4}$
$n_{2}$ = 0.892 $\times$ 0.0167 = 0.0148964 moles KMn$O_{4}$
$n_f$ = $n_1$ + $n_2$ = 0.058432 + 0.0148964 = 0.0733284 moles KMn$O_{4}$
$V_f$ = $V_1$ + $V_2$ = 35.2 + 16.7 = 51.9 mL = 0.0519 L KMn$O_{4}$
${C_{M}}_f$ = $\frac{0.0733284}{0.0519}$ $\approx$ 1.412 M
