A thin circular ring of radius 3.5 cm is resting horizontally on...

Given:
Radius of the ring, $R = 3.5\text{ cm} = 3.5 \times 10^{-2}\text{ m}$
Surface tension, $T = 0.05\text{ N m}^{-1}$

When a thin circular ring rests on a liquid surface, the liquid is in contact with both its inner and outer circumferences.
Total length of contact, $L = 2\pi R + 2\pi R = 4\pi R$

The excess upward force required to detach the ring is equal to the force due to surface tension.
$F = T \times L = T \times 4\pi R$

Substituting the values:
$F = 0.05 \times 4 \times \dfrac{22}{7} \times 3.5 \times 10^{-2}$
$F = 0.2 \times 22 \times 0.5 \times 10^{-2}$
$F = 2.2 \times 10^{-2}\text{ N}$
$F = 22 \times 10^{-3}\text{ N} = 22\text{ mN}$

Answer: $22\text{ mN}$