Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Blocks A and B are placed in contact, isolated, and allowed to come into equilibrium. Then they are placed in contact with block C. At that instant:a.TA = TB < TCb.TA = TB = TCc.TA = TB > TCd.TA + TB = TCe.TA - TB = TC

When put in contact each other, and isolated, both blocks will exchange heat till they reach to thermal equilibrium.
During this process, the body at a higher temperature, will loss heat, tat it will be gained by the other body.
The equilibrium condition will be reached when the following equation be met:

[tex]\Delta Q = c_{st}* m_{A} * (T_{fin} - T_{0A} ) = c_{st}* m_{B} * (T_{0B} - T_{fin} )[/tex]

Replacing by the values of T₀A = 300º C, and T₀B = 400ºC, and simplifying common terms as mA = mB, we can solve for Tfin, as follows:

[tex](400 \ºC - T_{fin}) = (T_{fin} - 300 \ºC) \\ \\ 2* T_{fin} = 700\ºC\\ \\ T_{fin} = 350\ºC[/tex]

So, when both blocks reach to equilibrium, they will be at a common final temperature, 350ºC.
When put in contact with block C, at the same temperature, at that instant, the three blocks will have the same common temperature of 350 ºC.
So, option b) is the right one.