The volume of $0.25 \mathrm{MNaOH}$ to be added to $250 \mathrm{mL}$ of $0.15 \mathrm{MNaOH}$ so that the resultant solution is $0.2 \mathrm{M}$ would be

. Let V be the volume of $0.25 \mathrm{MNaOH}$ solution. Total amount of $\text{NaOH}$ after mixing the two solutions is $n=V\left(0.25{\mathrm{molL}}^{-1}\right)+\left(0.25\mathrm{L}\right)\left(0.15{\mathrm{molL}}^{-1}\right)$

Total volume of the solution  $=V+0.25\mathrm{L}$

Molarity of the resultant solution  $M=\frac{V\left(0.25{\mathrm{molL}}^{-1}\right)+\left(0.25\mathrm{L}\right)\left(0.15{\mathrm{molL}}^{-1}\right)}{V+0.25\mathrm{L}}$

Equating this to 0.2 M, we get $M=\frac{V\left(0.25{\mathrm{molL}}^{-1}\right)+\left(0.25\mathrm{L}\right)\left(0.15{\mathrm{molL}}^{-1}\right)}{V+0.25\mathrm{L}}=0.2{\mathrm{molL}}^{-1}$

Solving for V, we get $V=0.25\mathrm{L}=250\mathrm{mL}$