From 1^st law of thermodynamics we know,

$$\Delta $$U = q + W

Option A :

In adiabatic process exchage of heat = 0

$$ \therefore $$ q = 0

$$ \therefore $$ From 1st law of thermodynaics, $$\Delta $$U = W

So option A is wrong.

Option B :

U is a state function. In cyclic process, initial state and final state both are same. So change in all the state function in cyclic process will be zero.

$$ \therefore $$ $$\Delta $$U = 0

$$ \therefore $$ From 1st law of thermodynaics,

q + W = 0 $$ \Rightarrow $$ q = -W

So option B is correct..

Option C :

In isochoric process volume (V) is constant. So dV = 0.

We know, W = $$ - \int {{P_{ex}}} dV$$

$$ \therefore $$ W = 0

$$ \therefore $$ From 1st law of thermodynaics,

$$\Delta $$U = q

So option C is correct.

Option D :

In isothermal process temerature (T) is constant. So dT = 0.

We know, $$\Delta $$U = nC_vdT

$$ \therefore $$ $$\Delta $$U = 0

$$ \therefore $$ From 1st law of thermodynaics,

q + W = 0 $$ \Rightarrow $$ q = -W

So option D is correct.