It is a non-traditional accumulative saving plan with $10,000 deposited semi-annually and compounded 
quarterly at 4% per annum. 

It means that the quarterly effective rate is  0.04/4 = 0.01 and the equivalent
semi-annual effective rate is  =  = 1.0201.


    +-------------------------------------------------------------------+
    |  So, this non-traditional accumulative saving plan is equivalent  |
    |   to the ordinary annuity with semi-annual deposits of $10,000    |
    |   and the semi-annual effective rate of compounding r = 1.0201.   |
    +-------------------------------------------------------------------+


Now use the general formula for a classic Ordinary Annuity saving plan


    FV = ,    (1)


where  FV is the future value of the annuity;  P is the semi-annual deposit; r is the semi-annual 
effective percentage yield presented as a decimal; n is the number of deposits.


Under the given conditions, P = 10000;  r = 0.0201.  So, according to (1), the formula for
the future value is


    FV = .


So, we should find n, the number of deposits (or the number of semi-annual periods)  from this equation

    50000 = .


Simplify it by dividing both sides by 10000

     = ,

or

    5 = .


Simplify it further, step by step

    5*0.0201 = ,

    0.1005 = ,

    0.1005 + 1 = ,

    1.1005 = .


Take logarithm base 10 of both sides

    log(1.1005) = n*log(1.0201)

and find n

    n =  = 4.81  (approximately).


Finally, round the decimal value of 4.81 to the closest GREATER integer value of 5 
in order for the bank be in position to complete the last semi-annual compounding.


At this point, the solution is complete.


The ANSWER is: 5 semi-annual periods are needed.