A person moved from A to B on a circular path as shown in figure....

The distance travelled by the person $d$ will be equal to the arch length.

$⇒ d = R\theta$

$⇒ 60 = R\left(\dfrac{3\pi}{4}\right)$

$⇒ R = \dfrac{60 \times 4}{3\pi} = \dfrac{80}{\pi}m$

Now the magnitude of displacement can be found using subtraction of vectors. Therefore,

$\left|s\right| = \sqrt{{{R}_{2}}^{2} + {{R}_{1}}^{2} - 2{R}_{1}{R}_{2}cos\theta} ⇒ \left|s\right| = \sqrt{{R}^{2} + {R}^{2} - 2{R}^{2}\cos 135 °}$

$⇒ \sqrt{2{R}^{2} - 2{R}^{2}\left(- 0 . 7\right)}$

$⇒ \sqrt{3 . 4{R}^{2}} = \sqrt{3 . 4{\left(\dfrac{80}{\pi}\right)}^{2}}$

$⇒ R = 47m$