Find out product (X)

The correct answer is:
$\text{Ph}-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_2-\text{OH}$

Let us break down the reaction sequence step-by-step:

Step 1: Hydroboration-Oxidation of Styrene
The starting material shown in the image is styrene:
$\text{Ph}-\text{CH}={\text{CH}}_2$
When treated with diborane in tetrahydrofuran (${\text{B}}_2{\text{H}}_6/\text{THF}$) followed by alkaline hydrogen peroxide (${\text{H}}_2{\text{O}}_2/{\text{OH}}^{-}$), the alkene undergoes hydroboration-oxidation. This reaction is an anti-Markovnikov addition of water across the double bond, where the hydroxyl group ($-\text{OH}$) attaches to the less substituted terminal carbon. This yields 2-phenylethan-1-ol:
$\text{Ph}-{\text{CH}}_2-{\text{CH}}_2-\text{OH}$

Step 2: Substitution with HBr
The alcohol 2-phenylethan-1-ol is then treated with hydrobromic acid ($\text{HBr}$). The hydroxyl group is protonated to form water (a good leaving group), which is substituted by the bromide ion (${\text{Br}}^{-}$). This converts the alcohol into an alkyl halide, 2-phenylethyl bromide:
$\text{Ph}-{\text{CH}}_2-{\text{CH}}_2-\text{Br}$

Step 3: Grignard Reagent Formation
The alkyl halide is reacted with metallic magnesium ($\text{Mg}$) in dry ether. The magnesium inserts into the carbon-halogen bond to form a nucleophilic Grignard reagent, 2-phenylethylmagnesium bromide:
$\text{Ph}-{\text{CH}}_2-{\text{CH}}_2-\text{MgBr}$

Step 4: Reaction with Formaldehyde (HCHO)
The nucleophilic carbon of the Grignard reagent attacks the electrophilic carbonyl carbon of formaldehyde ($\text{HCHO}$). This nucleophilic addition adds one carbon atom to the chain and forms an alkoxide intermediate:
$\text{Ph}-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_2-{\text{O}}^{-}{\text{MgBr}}^{+}$
Subsequent acidic workup protonates the alkoxide, yielding the primary alcohol product (X), 3-phenylpropan-1-ol:
$\text{Ph}-{\text{CH}}_2-{\text{CH}}_2-{\text{CH}}_2-\text{OH}$