It's given that: $3^{2019}$ is divided by 10.

Now, $3^1 = 3$ $3^2 = 9$ $3^3 = 27$ $3^4 = 81$ $3^5 = 243$ $3^6 = 729$

Since, unit place of the power of 3 repeats after every 4 steps (i.e. it has a cyclicity of 4). Now, on dividing 2019 by 4 we get a remainder of 3.

Hence, $3^{2019}$ will have the same last digit as that of $3^3$, i.e. 7. $(3^3)/10 = 27/10$

Hence, the remainder will be 7.