Determine surface area of heat exchange tube
The outlet temperature  should be 53.3.  The log-mean temperaturr difference should be positive.  Calculate the heat load and divide by the log-mean temperature difference and by the heat transfer coefficient to get the area.

ADDENDUM

This addendum is for the benefit of @MaxW (the downvoter).


The heat load $\dot{Q}$ of the heat exchanger is given by:  $$\dot{Q}=\dot{m}_HC(T_{H,in}-T_{H,out})=20(4.183)(80-60)=1673\ kW$$This heat load represents the rate of heat transfer from the hot fluid to the cold fluid.
The exit temperature of the cold stream is determined from $$\dot{m}_CC(T_{C,out}-T_{C,in})=12(4.183)(T_{C,out}-20)=\dot{Q}=1673$$


Solving for the exit temperature of the cold stream gives  $$T_{C,out}=53.33\ C$$
3.  The log-mean temperature difference is given by:
$$(\Delta T)_{LM}=\frac{(T_{H,in}-T_{C,out})-(T_{H,out}-T_{C,in})}{\ln{[(T_{H,in}-T_{C,out})/(T_{H,out}-T_{C,in})]}}=\frac{26.67-40}{\ln{[26.67/40]}}=32.88\ C$$


The heat transfer area A is determined from the equation:  $$U_oA(\Delta T)_{LM}=2.0A(32.88)=\dot{Q}=1673$$Solving this equation for the heat transfer area A gives:
$$A=25.4\ m^2$$


Questions?