Answer
$4~kg/m$
Work Step by Step
Let $\lambda$ be the linear density of the wire at any point $x$.
Then:
$mass = \int_{0}^{x} \lambda ~dx = x(1+\sqrt{x})~kg$
Then:
$\lambda = \frac{d}{dx}[x(1+\sqrt{x})]$
$\lambda = \frac{d}{dx}(x+x^{3/2})$
$\lambda = 1+\frac{3}{2}\sqrt{x}$
When $x = 4~m$:
$\lambda = 1+\frac{3}{2}\sqrt{4}$
$\lambda = 1+3$
$\lambda = 4~kg/m$
