[Tgx]

I know that 1 mole of gas at 0°C and 1 atm has a volume of 22.4 liters. So first, I'll convert the 21 liters of Helium at 1.15 atmospheres to x liters of Helium at 1 atm. For this, I will use the expression:
$$P_{1}V_{1}=P_{2}V_{2}$$
$$1.15*21=1*V_{2}$$
$$V_{2}=24.15$$
So, at 1 atm, we have 24.15 liters of gas. We can use a comparison:
$$\frac{n}{N_{a}}=\frac{24.15}{22.4}$$
Where n is the number of atoms for 24.15 liters of gas and N_a is the number of atoms for one mole (22.4 liters) of gas.
$$n=\frac{24.15}{22.4}*N_{a}$$
$$n≈6.49*10^{23}$$

Is this correct? Thank you in advance.

[Borek]

OK, but a bit overcomplicated. Just plug the numbers into pV=nRT, find number of moles and you are done.

[billnotgatez]

OK, but a bit overcomplicated. Just plug the numbers into pV=nRT, find number of moles and you are done.

Does that mean that the value for R has to be provided, looked up, or memorized (with proper units).
I guess we can assume the student knows the conversion from kelvin to Celsius.

I do note that the OP has already known
  the relationship of volume and moles at 0 C and 1 atm
  the relationship of moles and  atoms of helium.

[Borek]

Does that mean that the value for R has to be provided, looked up, or memorized (with proper units).

Yes, but this is one of a few universal constants that are used so often they are worth memorizing (and let's not forget that remembering R is much more universal than remembering volume of 1 mole of gas at STP ).