Step 1: Use Bohr's quantization of angular momentum.
The angular momentum of an electron in the $n$th orbit is quantized as \[ L = \frac{nh}{2\pi} \] where $h$ is Planck's constant.
Step 2: Insert the numbers to find $n_x$.
With $L = 1.051 \times 10^{-34}$, $h = 6.6 \times 10^{-34}$, and $\pi = 3.14$, \[ 1.051 \times 10^{-34} = \frac{n_x (6.6 \times 10^{-34})}{2(3.14)} \]
Step 3: Simplify the constant.
Since $\frac{6.6}{6.28} \approx 1.051$, the equation becomes $1.051 = 1.051\,n_x$, giving $n_x = 1$. The electron sits in the ground state.
Step 4: Write the excitation energy formula.
Moving from $n = 1$ to $n = 2$ needs \[ \Delta E = 2.18 \times 10^{-18}\left(\frac{1}{1^2} - \frac{1}{2^2}\right) \]
Step 5: Evaluate the bracket.
\[ 1 - \frac{1}{4} = \frac{3}{4} \]
Step 6: Finish the calculation.
So $\Delta E = 2.18 \times 10^{-18} \times \frac{3}{4} = 1.635 \times 10^{-18}$ J, which is option 4.
\[ \boxed{1.635 \times 10^{-18}\ \text{J}} \]