Three inductances are connected as shown in the figure. The equivalent inductance between $A$ and $B$ is

Equivalent inductance between points A and B

The circuit contains two 0.7 H inductors in parallel, which are then in series with a 0.85 H inductor.

For parallel inductors, the equivalent is $L_p = \frac{L_1 \cdot L_2}{L_1 + L_2} = \frac{0.7 \cdot 0.7}{0.7 + 0.7} = \frac{0.49}{1.4} = 0.35$ H.

The total inductance is then $L_{eq} = L_p + L_3 = 0.35 + 0.85 = 1.20$ H.

This matches option B of $1.20$ H.