Inverse Fourier transform Of a triangular impulse
It's nice to see alternative paths, like this one: First of all, properties:


Frequency shifting: $x(t)e^{j2\pi f_0 t} \longleftrightarrow X(f-f_0)$ 
Duality: if $x(t) \longleftrightarrow X(f)$ then $X(t) \longleftrightarrow x(-f)$


You have correctly expressed the depicted transform: $$X(f) = \frac{1}{2}\mathrm{tri}\Big(\frac{f+f_0}{B}\Big) - \frac{1}{2}\mathrm{tri}\Big(\frac{f-f_0}{B}\Big)$$

As you said, $\mathrm{tri}\Big(\frac{t}{B}\Big) \longleftrightarrow B\mathrm{sinc}^2(fB)$. From the duality property, you have $$B\mathrm{sinc}^2(tB) \longleftrightarrow \mathrm{tri}\Big(\frac{-f}{B}\Big)$$ but since $\mathrm{tri(\cdot )}$ is an even function, you can write $$B\mathrm{sinc}^2(tB) \longleftrightarrow \mathrm{tri}\Big(\frac{f}{B}\Big)$$

Now you have your FT-pair you need. If you apply the frqeuency shifting property on $\mathrm{tri}\Big(\frac{f\pm f_0}{B}\Big)$, you can easily get... what? :-) You can continue from this point.