1. For an electron in an atomic orbital, the magnitude of orbital angular momentum is given by $L = \sqrt{l(l+1)}\,\frac{h}{2\pi}$ where $l$ is the azimuthal quantum number.

2. For a $3s$ orbital, the subshell is $s$, so $l=0$

3. Substituting into the formula: $L = \sqrt{0(0+1)}\,\frac{h}{2\pi} = 0\cdot \frac{h}{2\pi}=0$

4. Given that the orbital angular momentum is written as $\frac{x h}{2\pi}$ we compare: $\frac{x h}{2\pi}=0$ Hence, $x=0$

Therefore, the nearest integer value of $x$ is $0$.