Calculate for the VTH:

$\text{Apply KVL to mesh I} \\ (10+50)I_1-10I_2-50I_2=80 \\ 60I_1-60I_2=80..............(I) \\ \text{Apply KVL to mesh II} \\ (50+20+50)I_2-50I_1-10I_1=0 \\ -60I_1+120I_2=0..............(II) \\ \text{Solving (I) &(II) we get} \\ I1=2.666A \hspace{0.5cm} I2=1.333A \\ \text{Writing KVL equation for VTH from} \\ -10I_2+50(I_1-I_2)-V_{TH}=0 \\ V_{TH}=53.32 V$

Calculation for RTH:-

$RTH=8.333+8.333=16.666 \Omega$

The venin’s circuits is

$I_L$ $=\dfrac{V_{TH}}{R_{TH}+R_L} \\ =\dfrac{53.32}{16.666+60} \\ =\dfrac{53.32}{76.666}$

$I_L=0.6955A$