The number of sp$$^2$$ hybridised carbons present in "Aspartame" is

First we recall the definition of an $$\text{sp}^2$$-hybridised carbon. A carbon atom is $$\text{sp}^2$$-hybridised when it forms three $$\sigma$$-bonds and one unhybridised $$p$$-orbital participates in a $$\pi$$-bond; this happens in every carbon that is part of a double bond ($$C=C$$ or $$C=O$$) and in every carbon atom of an aromatic ring such as benzene.

Now we write the full structural formula of Aspartame (methyl L-α-aspartyl-L-phenylalaninate). In condensed form the molecule can be expressed as

$$$\mathrm{CH_3OOC\!-\!CH(NH_2)\!-\!CH_2\!-\!CO\!-\!NH\!-\!CH(CH_2\!-\!C_6H_5)\!-\!COOCH_3}$$$

For clarity we expand it, separating every carbon so that we can inspect their hybridisation:

$$$ \begin{aligned} &\text{(i) } \mathrm{COOCH_3} &&\longrightarrow &&\underline{\mathrm{O=}}\boxed{\mathrm{C}}\!-\!\mathrm{O}\!-\!\mathrm{CH_3} \\ &\text{(ii) } \mathrm{-CH(NH_2)-} &&\longrightarrow &&\boxed{\mathrm{C}}\!-\!\mathrm{H},\;\mathrm{NH_2} \\ &\text{(iii) } \mathrm{-CH_2-} &&\longrightarrow &&\boxed{\mathrm{C}}\!-\!\mathrm{H_2} \\ &\text{(iv) } \mathrm{-CO-} &&\longrightarrow &&\underline{\mathrm{O=}}\boxed{\mathrm{C}}\!-\!\mathrm{NH-} \\ &\text{(v) } \mathrm{-CH(}\mathrm{CH_2\!-\!C_6H_5}\mathrm{)-} &&\longrightarrow &&\boxed{\mathrm{C}}\!-\!\mathrm{H},\;\mathrm{CH_2},\;\mathrm{C_6H_5} \\ &\text{(vi) } \mathrm{-COOCH_3} &&\longrightarrow &&\underline{\mathrm{O=}}\boxed{\mathrm{C}}\!-\!\mathrm{O}\!-\!\mathrm{CH_3} \end{aligned} $$$

The phenyl ring $$\mathrm{C_6H_5}$$ is an ordinary benzene ring, so every one of its six carbons is $$\text{sp}^2$$-hybridised.

We now count the $$\text{sp}^2$$ carbons one by one.

1. In the benzene (phenyl) ring there are $$6$$ carbons. All $$6$$ are $$\text{sp}^2$$.

2. Each carbonyl group $$\mathrm{C=O}$$ contributes one $$\text{sp}^2$$ carbon because the carbon in $$\mathrm{C=O}$$ is involved in one double bond. Aspartame possesses three separate carbonyls:   (a) the ester carbonyl on the extreme left,   (b) the amide carbonyl in the middle, and   (c) the second ester carbonyl on the right. Therefore the number of $$\text{sp}^2$$ carbons from carbonyls is $$3$$.

3. All remaining carbons (those in $$\mathrm{CH_3}$$, $$\mathrm{CH_2}$$ and $$\mathrm{CH}$$ groups that are not part of a double bond or an aromatic ring) are $$\text{sp}^3$$, so they do not contribute to our count.

Adding the contributions, we have

$$$\text{Total sp}^2 \text{ carbons} \;=\; 6 \;(\text{benzene}) \;+\; 3 \;(\text{carbonyls}) \;=\; 9.$$$

So, the answer is $$9$$.