1.4137449e-18, as others have said, is incredibly small.
The equation given allows you to calculate the capacitor required for an LC (inductor-capacitor) circuit resonant at the given frequency. The number you calculated is for an L of 1 Henry. Looking at the equation it's clear that as the inductance (L) decreases, the capacitance increases. Inductors can be found in a range of inductances from perhaps 1nH and larger. Given that a nanoHenry is \$0.000000001\$ Henries (\$1 \times 10^{-9}\$), you should be able to see that the capacitance required could be much more sensible.
Picking an inductor value of \$100\$nH (or \$0.0000001\$ Henries) gives \$1.4137 \times 10^{-11}\$ Farads, or \$14\$ pico Farads of capacitance. By reducing the inductor size, you can increase the capacitor value further - an inductor of \$10\$nH would give a capacitance of \$141\$pF and so on.
If you don't understand scientific notation it's probably worth finding a maths textbook which explains it, because you can't really do any engineering or electronics design without having a good grasp of maths.
E
is being divided by 10 followed by 18 zeroes? So the actual number is 0.00000000000000000147726744996646, which is practically zero. \$\endgroup\$