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**Hint:**To solve this question, we will first check the possibility of A value. Then putting the possible value of A in the above equation we will obtain the value of B. After that we will check the error. We can also do this by an alternative method which we will discuss in note.

**Complete step-by-step answer:**

The given equation is $BA \times B3 = 57A$

Here A is in the unit place of the product side or we can say in the right hand side which means when we will multiply BA with B3, the product of multiplication of the unit place of both the terms of the left hand side must contain A in its unit place.

That means when we will multiply 3 with A, their product value must contain A in unit place.

If we will check the possibility, then two possibilities arrive. i.e. either A=0 or A=5 .

$BA \times B3 = 57A$

As B is in the 10’s place, then expanding the equation we get,

$\left( {10 \times B + A} \right) \times \left( {10 \times B + 3} \right) = \left( {57 \times 10} \right) + A$

$ \Rightarrow \left( {10B + A} \right) \times \left( {10B + 3} \right) = 570 + A$

$ \Rightarrow 100{B^2} + 10(A \times B) + 30B + 3A = 570 + A$

Subtracting A from both sides we get,

$100{B^2} + 10(A \times B) + 30B + 2A = 570$……………..(1)

For A=0

Putting A=0 in equation 1 we get,

$100{B^2} + 30B = 570$

Dividing 10 on each side we get,

$10{B^2} + 3B = 57$………..(2)

No B satisfies this as

$3B = 7$

For A=5

Putting A=5 in equation 1 we get,

$100{B^2} + 80B + 10 = 570$

Subtracting 10 from both side and then dividing both side with 10 we get,

$10{B^2} + 8B = 56$…………….(3)

$ \Rightarrow 5{B^2} + 4B = 28 = 20 + 8$

Hence B=2

Hence the positional value of B and A is 2, 5 respectively.

**So, the correct answer is “Option D”.**

**Note:**Checking error : $25 \times 23 = 575$. Error checked.

For step-2, you have to check it by multiplying 3 with 0 to 1.

You can use middle term factorization method or can put the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ to get the value of B in equation 2 and 3.

You might mistaken the terms of equation in the question i.e. AB is multiplication of A and B, B3 as 3 multiplied with B and 57A as multiplication of 57 and A.

You can also solve it using an alternative method, by checking through options. ( if options are given)

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