To use linearization you need to identify an appropriate function and a point to linearize that function.
The function $f(x) = x^{1/3}$ seems appropriate, and you can evaluate $f(64)$ directly. The linearization of $f$ at $a = 64$ is given by $$y = f(64) + f'(64)(x-64).$$ Since $f'(x) = \dfrac 13 x^{-2/3}$ you have $f(64) = 4$ and $f'(64) = \dfrac 1{3 \cdot 16}$ so that the approximation with $x=61$ is $$y = 4 + \dfrac 1{3 \cdot 16} (61-64) = 4 - \frac 1{16} = 3.9375.$$
The actual value is pointed out in the comments, so the linearization doesn't give you enough decimal places. Since you are in Calc 2 you have probably seen Taylor's theorem, so instead you should try the second order approximation $$y = f(64) + f'(64)(x-64) + \frac{f''(64)}{2}(x-64)^2$$ and see what that yields.