1. Interpret the experiment

Let the two randomly chosen real numbers be $x$ and $y$, where $x,y \in [0,60].$

Since both are chosen uniformly and independently, the sample space is the square $0 \le x \le 60, \quad 0 \le y \le 60,$ with total area $60 \times 60 = 3600.$

The event $A$ is $|x-y| \le a.$ So we need the probability that the point $(x,y)$ lies within the strip between the lines $y=x+a \quad \text{and} \quad y=x-a.$

2. Use area method

Inside the square $[0,60]\times[0,60]$, the region where $|x-y|>a$ consists of two congruent triangles:

• one above the line $y=x+a$,
• one below the line $y=x-a$.

Each triangle has side length $60-a,$ so area of each triangle is $\frac{1}{2}(60-a)^2.$

Hence total area excluded from the strip is $2\cdot \frac{1}{2}(60-a)^2=(60-a)^2.$

Therefore favorable area is $3600-(60-a)^2.$

So $P(A)=\frac{3600-(60-a)^2}{3600}.$

Given $\frac{3600-(60-a)^2}{3600}=\frac{11}{36}.$

3. Solve for $a$

Multiply both sides by $3600$: $3600-(60-a)^2=3600\cdot \frac{11}{36}=1100.$

Thus $(60-a)^2=3600-1100=2500.$

So $60-a=\pm 50.$

Since $0\le a\le 60$, we must have $60-a\ge 0$, hence $60-a=50.$

Therefore $a=10.$

4. Final answer

$\boxed{10}$

5. Comparison with stored answer

Stored correct answer = $10$.

This matches the derived answer.