Circuit analysis using Mesh current method
This is how I solve it.
First I marked the mesh currents

And due to the fact that in mesh one we have a constant current source, we know that:
$$I_1 = 0.5mA$$
Thus the equation for \$I_2\$ mesh will look like this:
$$3(I_2 + I_3 + 0.5) + 10I_2 + 1.5(I_2 + I_3)=0$$
and for \$I_3\$ we have:
$$-8 + 2I_3 + 1.5(I_3+I_2)+3(I_3 + I_2 + 0.5)=0$$
And the solution is:
here
$$I_1 = 0.5mA$$
$$I_2 =-0.527027mA$$
$$I_3 = 1.36486mA$$
And the voltage across \$R_3\$ resistor is equal to:
$$V_{R3} = (I_1 + I_2 + I_3)\times 3k\Omega = 4.01351V$$