A body is projected vertically from earth's surface with $\left(\frac{1}{3}\right)^{\text {rd }}$ of escape velocity. The maximum height reached by the body is ( $\mathrm{R}=$ radius of earth)

Given that the escape velocity from Earth's surface is $v_e = \sqrt{\frac{2GM}{R}}$, with $G$ the gravitational constant, $M$ the Earth's mass, and $R$ its radius, the body is projected upward with velocity $v = \frac{1}{3}v_e$.

Using conservation of mechanical energy, the initial energy at the surface equals the final energy at maximum height $h$ where velocity is zero:

$\frac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}$

Substituting $v = \frac{1}{3}\sqrt{\frac{2GM}{R}}$:

$\frac{1}{2}m \left(\frac{1}{9} \cdot \frac{2GM}{R}\right) - \frac{GMm}{R} = -\frac{GMm}{R+h}$

After simplification, this reduces to:

$\frac{GMm}{9R} - \frac{GMm}{R} = -\frac{GMm}{R+h}$

Dividing through by $GMm$ yields:

$\frac{1}{9R} - \frac{1}{R} = -\frac{1}{R+h}$

Combining the left side gives:

$-\frac{8}{9R} = -\frac{1}{R+h}$

Multiplying both sides by $-1$ and cross-multiplying results in:

$8(R + h) = 9R$

Solving for $h$:

$h = \frac{R}{8}$

Maximum height reached is $\frac{R}{8}$, corresponding to option B.