Step 1: Identify the reaction.
Borax, $Na_2B_4O_7$, reacts with water (hydrolysis) to give an acid plus an alkali. We need the central atom of the acid.
Step 2: Write the products.
The hydrolysis gives orthoboric acid and sodium hydroxide, \[ Na_2B_4O_7 + 7H_2O \rightarrow 4H_3BO_3 + 2NaOH \] So the acid is $H_3BO_3$ and the alkali is NaOH.
Step 3: Spot the central atom.
In $H_3BO_3$ the central atom is boron, surrounded by oxygen and hydrogen.
Step 4: Recall boron's behaviour in water.
Boric acid is a weak monobasic Lewis acid. It does not give up a proton directly; instead it accepts a hydroxide from water to form the borate ion $[B(OH)_4]^-$.
Step 5: Find the hybridisation of that species.
In $[B(OH)_4]^-$ boron is bonded to four groups, so it uses four equivalent orbitals, which means $sp^3$ hybridisation.
Step 6: State the answer.
The central boron, in the form relevant to its acidic behaviour, is $sp^3$ hybridised, matching option 1.
\[ \boxed{sp^3} \]