Вернуться .файл XLSX в Excel с веб-API

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Error	CS1503	Argument 1: cannot convert from 'System.Collections.Generic.List<WebApi2.Controllers.Record>' to 'System.IO.Stream'	

[HttpGet]
        [Route("ExcelDownload")]
        public HttpResponseMessage ExcelDownload()
        {                       
            var reportStream = RecordInfo();
            var result = Request.CreateResponse(HttpStatusCode.OK);
            result.Content = new StreamContent(reportStream); // Getting error here
            result.Content.Headers.ContentType = new MediaTypeHeaderValue("application/octet-stream");
            result.Content.Headers.ContentDisposition = new ContentDispositionHeaderValue("attachment")
            {
                FileName = "Report.xlsx"
            };

            return result;
        }


        public List<Record> RecordInfo()
        {
            List<Record> recordobj = new List<Record>();
            recordobj.Add(new Record { FName = "Smith", LName = "Singh", Address = "Knpur" });
            recordobj.Add(new Record { FName = "John", LName = "Kumar", Address = "Lucknow" });
            recordobj.Add(new Record { FName = "Vikram", LName = "Kapoor", Address = "Delhi" });
            recordobj.Add(new Record { FName = "Tanya", LName = "Shrma", Address = "Banaras" });
            recordobj.Add(new Record { FName = "Malini", LName = "Ahuja", Address = "Gujrat" });
            recordobj.Add(new Record { FName = "Varun", LName = "Katiyar", Address = "Rajasthan" });
            recordobj.Add(new Record { FName = "Arun  ", LName = "Singh", Address = "Jaipur" });
            recordobj.Add(new Record { FName = "Ram", LName = "Kapoor", Address = "Panjab" });
            recordobj.Add(new Record { FName = "Vishakha", LName = "Singh", Address = "Banglor" });
            recordobj.Add(new Record { FName = "Tarun", LName = "Singh", Address = "Kannauj" });
            recordobj.Add(new Record { FName = "Mayank", LName = "Dubey", Address = "Farrukhabad" });
            return recordobj;
        }