Step 1: Recall the Lyman series condition.
A Lyman line is emitted when an electron falls to the ground level $n = 1$. The energy released coming from level $n_x$ is \[ \Delta E = 2.18 \times 10^{-18}\left(1 - \frac{1}{n_x^2}\right)\ \text{J} \]
Step 2: Plug in the given emission energy.
We are told this energy equals $1.635 \times 10^{-18}$ J, so \[ 1.635 \times 10^{-18} = 2.18 \times 10^{-18}\left(1 - \frac{1}{n_x^2}\right) \]
Step 3: Solve for $n_x$.
Dividing both sides by $2.18 \times 10^{-18}$ gives $0.75 = 1 - \frac{1}{n_x^2}$, hence $\frac{1}{n_x^2} = 0.25 = \frac{1}{4}$, so $n_x = 2$.
Step 4: Set up the required excitation.
Now we need the energy to lift the electron from $n = 2$ to $n = 3$, \[ \Delta E = 2.18 \times 10^{-18}\left(\frac{1}{2^2} - \frac{1}{3^2}\right) \]
Step 5: Simplify the bracket.
\[ \frac{1}{4} - \frac{1}{9} = \frac{9 - 4}{36} = \frac{5}{36} \]
Step 6: Compute the final energy.
So $\Delta E = 2.18 \times 10^{-18} \times \frac{5}{36} \approx 3.03 \times 10^{-19}$ J, which rounds to option 1.
\[ \boxed{\Delta E \approx 3 \times 10^{-19}\ \text{J}} \]