**Question Analysis:**

We are given the accelerations of two particles, A and B. Particle A has a constant acceleration of 3 m/s² in a certain direction, while particle B has a constant acceleration of 4 m/s² towards particle A. We need to find the value of acceleration a in SI units, given that the time after which particle B meets particle A is 2lb b² a.

**Solution:**

To solve this problem, we can use the equations of motion to find the time it takes for particle B to meet particle A. Let's break down the solution into steps:

**Step 1: Finding the time it takes for particle B to meet particle A:**

We can use the equation of motion for particle A to find its displacement. Since particle A starts from rest, the equation becomes:

s_A = 0.5 * a * t²

where s_A is the displacement of particle A and t is the time taken. We know that the acceleration of particle A is given as a = 3 m/s².

Next, we can use the equation of motion for particle B to find its displacement. Since particle B starts from rest and has a constant acceleration of 4 m/s² towards particle A, the equation becomes:

s_B = 0.5 * b * t²

where s_B is the displacement of particle B and t is the time taken. We know that the acceleration of particle B is given as b = 4 m/s².

Since both particles start from rest, the total displacement of particle A and particle B should be equal when they meet. Therefore, we can equate their displacements:

0.5 * a * t² = 0.5 * b * t²

**Step 2: Solving for time t:**

To solve for time t, we can cancel out the common terms and rearrange the equation:

a * t² = b * t²

Dividing both sides by t², we get:

a = b

Therefore, the acceleration a is equal to the acceleration b.

**Step 3: Finding the value of acceleration a in SI units:**

From step 2, we know that the acceleration a is equal to the acceleration b, which is given as 4 m/s². Therefore, the value of acceleration a in SI units is 4 m/s².

**Answer:**

The value of acceleration a in SI units is 4 m/s².