Thanks for providing more information about the relay. It's now understood the complete part number is HK19F-DC5V-SHG. It's not always necessary to have such a complete number like this. But in this case the company appears to make four different sub-versions of their \$5\:\text{V}\$ version. I don't see that very often. But there it was.
Here is the HK19F-DC5V-SHG datasheet. You can see that they specify four different versions at \$5\:\text{V}\$.

So you look at the ordering information to clarify, given your fuller part number:

This means you have the \$200\:\text{mW}\$ version, which is \$125 \:\Omega\$, nominal. The equilibrium full current will then be \$\sqrt{\frac{200\:\text{mW}}{125 \:\Omega}}=40\:\text{mA}=\frac{5\:\text{V}}{125 \:\Omega}\$.
Now you know the maximum collector current for the BJT. The rule of thumb is to specify \$\frac1{10}\$th of that for the base current. But that rule isn't always good. In the case of power transistors like the 2N3055, twice as much or more may be required. But you won't need one of those, here.
Technically, the BJT datasheet should always be consulted. But in this case, with only \$40\:\text{mA}\$ that can be handled well by most general purpose small signal BJTs used for switching purposes, it's not necessary. There's a different question that is more important.
What's the I/O pin current compliance?
I don't have a datasheet to look at for your device. You didn't provide a way to directly find it and there are too many to sort through to see if there is a common value available or if they differ so much that nothing can be said. So this will be the uncertainty.
But given the rule of thumb, a base current of \$\frac1{10}\$th of \$40\:\text{mA}\$ or \$4\:\text{mA}\$, I'd say that unless the circumstances are unusual it's likely okay.
At \$40\:\text{mA}\$ for the collector current, the base voltage will likely be in the area of \$700\:\text{mV}+28\:\text{mV}\cdot\ln\left(\frac{40\:\text{mA}}{1\:\text{mA}}\right)\approx 800\:\text{mV}\$. (I didn't add all my thinking here, but I'm allowing for some warmth in the BJT and taking some experience with small signal BJTs into the bargain.)
I'd also expect to see no more than \$100\:\Omega\cdot 4\:\text{mA}=400\:\text{mV}\$ drop at the I/O pin (it will likely be less, but this is a probable worst case.
So the base resistor value should be \$\frac{3.3-400\:\text{mV}-800\:\text{mV}}{4\:\text{mA}}=525 \:\Omega\$. Since I was conservative up to this point (the \$\frac1{10}\$th rule of thumb, the output impedance of the I/O pin, etc), a larger value would be the direction to head. So either \$560\:\Omega\$ or else \$680\:\Omega\$. Both would likely work very well. (Still higher values may also work quite well. But unless you have a reason, such as not having those values handy, then being on the conservative side will allow most any 'junkbox' small signal BJT to be used here.)