Let us sketch the diagram for the motion of the particle of mass $m$ along the circle of radius $R$ and indicate $x$ and $y$ axis, as shown in the figure.
(a) For the particle, change in momentum $\Delta \vec{p}=m v(-\vec{i})-m v(\vec{j})$
so, $|\Delta \vec{p}|=\sqrt{2} m v$
and time taken in describing quarter of the circle,
Hence, $\langle\vec{F}\rangle=\frac{|\Delta \vec{p}|}{\Delta t}=\frac{\sqrt{2} m v}{\pi R / 2 v}=\frac{2 \sqrt{2} m v^{2}}{\pi R}$
(b) In this case $\overrightarrow{p_{i}}=0$ and $\overrightarrow{p_{f}}=m w_{t} t(-\vec{i})$
so $|\Delta \vec{p}|=m w_{t} t$
Hence, $|<\vec{F}\rangle \mid=\frac{|\Delta \vec{p}|}{t}=m w_{t}$