$x = y^3$

$y = x^{1/3}$

$y' = \frac{1}{3}x^{-2/3}$

$y'' = -\frac{2}{9}x^{-5/3}$

At (1, 1)

$y' = \frac{1}{3}(1^{-2/3}) = \frac{1}{3}$

$y'' = -\frac{2}{9}(1^{-5/3}) = -\frac{2}{9}$

Radius of Curvature formula

$\rho = \dfrac{\left[ 1 + (y')^2 \right]^{3/2}}{\left\lvert y'' \right\lvert}$

$\rho = \dfrac{\left[ 1 + \left( \frac{1}{3} \right)^2 \right]^{3/2}}{\left\lvert -\frac{2}{9} \right\lvert}$

$\rho = 5.27 ~ \text{units}$ ← *answer*